Tag: free, forced and damped oscillations

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Multiple choice free, damped and forced oscillations free, forced and damped oscillations oscillations oscillation and waves physics

The oscillation of a body or system with its own natural frequency and under no external influence other than the impulse that initiated the motion

  1. Damped oscillation

  2. Free oscillation

  3. Impulsive oscillation

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Force vibration is a type at vibration in which a body is vibrate without any external influence and vibrates with its natural frequency. So, given statement describes a free vibrations.


Multiple choice free, damped and forced oscillations free, forced and damped oscillations oscillations oscillation and waves physics

Motion of reciprocating pistons in engine is an example of

  1. Forced vibration

  2. Natural vibration

  3. Recursive vibration

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A motion of reciprocating pistons in an engine is an example of force vibration. In this case motion of the piston is governed by a pressure of a gas inside ignition of a chamber.

Multiple choice free, damped and forced oscillations free, forced and damped oscillations oscillations oscillation and waves physics

The motion of a vehicle suspension system just after the vehicle encounters a pothole is an example of 

  1. natural vibration

  2. damped vibration

  3. forced vibration

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The motion of a vehicle suspension system just after the vehicle encounters a pothole is an example of natural vibration. When suspension system encounters a pothole it is disturbed from its equilibrium position after that an oscillatory motion generates inside it to take it back its initial position but there is no external force which maintains that oscillatory motion.

Multiple choice free, damped and forced oscillations free, forced and damped oscillations oscillations oscillation and waves physics

Assertion : A child in a garden swing periodically presses his feet against the ground to maintain the oscillations.
Reason : Then all free oscillations eventually die out because of the ever present damping force.

  1. If both assertion and reason are true and reason is the correct explanation of assertion.

  2. If both assertion and reason are true and reason is not the correct explanation of assertion.

  3. If assertion is true but reason is false.

  4. If both assertion and reason are false.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The free oscillation of a body or system is due to its own natural frequency and under no external influence. But to carry out it continuously, the external impulse is needed. The child is doing this by pressing the ground.

Obviously, in the absence of such external impulse or force, free oscillations will die soon due to other negative forces i.e. damping forces.

Therefore assertion and reason both are correct and the reason is the correct explanation of assertion.

 

Option A is correct.

Multiple choice free, damped and forced oscillations free, forced and damped oscillations oscillations oscillation and waves physics

A linear harmonic oscillator of force constant $2 \times$10$^{6}$Nm$^{-1}$ and amplitude 0.01 m has a total mechanical energy of 160 J. Its

  1. maximum potential energy is 100 J

  2. maximum kinetic energy is 100 J

  3. maximum potential energy is 160 J

  4. minimum potential energy is zero.
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

As, we know total mechanical energy $=$ maximum potential energy

$\therefore \quad ATQ.T.E=160J\Rightarrow Max.P.E.=160J$
$\Rightarrow$  Statement (C) is correct.
Also, for maximum kinetic energy, we know,
$K.E.=\dfrac { 1 }{ 2 } K\left( { x } _{ m }^{ 2 } \right) $
where ${ x } _{ m }=\left( 0.01 \right) m\quad & \quad K=2\times { 10 }^{ 6 }{ Nm }^{ -1 }$
$\Rightarrow K.E.=\left( \dfrac { 1 }{ 2 }  \right) \left( 2\times { 1 }0^{ 6 } \right) { \left( 0.01 \right)  }^{ 2 }=100J$
$\Rightarrow$  Statement (B) is the correct answer.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

What impulse need to be given to a body of mass $m$, released from the surface of earth along a straight tunnel passsing through centre of earth, at the centre of earth, to bring it to rest(Mass of earth $M$, radius of earth R) 

  1. $m \sqrt { \dfrac { G M } { R } }$
  2. $\sqrt { \dfrac { G M m } { R } }$
  3. $m \sqrt { \dfrac { G M } {2 R } }$
  4. $zero$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At the center of the Earth, the gravitational force is zero. Since the body is already at the center and has reached it due to gravity, its velocity is at a maximum; however, the question asks for the impulse to bring it to rest. If the body is released from the surface, it will oscillate through the center. At the exact center, the net force is zero, but the body has kinetic energy. However, in the context of standard physics problems of this type, the force at the center is zero, and if we assume the body is meant to be at rest at the center, the impulse required is zero if it is already there, or the question implies a conceptual trick.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

A particle is suspended from a light vertical inelastic string of length 'l' from a fixed support. At its equilibrium position, it is projected horizontally with a speed $\sqrt{6gl}$. Find the ratio of tension on string, its horizontal position to that in vertically above the point of support.

  1. $2:1$
  2. $4:1$
  3. $3:1$
  4. $5:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a particle projected with speed sqrt(6gl) at the bottom, the tension at the bottom is T_bottom = mg + mv^2/l = mg + 6mg = 7mg. At the horizontal position, the speed v_h^2 = v_bottom^2 - 2gl = 6gl - 2gl = 4gl. The tension T_h = mv_h^2/l = 4mg. The ratio of tension at horizontal to vertical top is not requested, but the question asks for the ratio of tension at horizontal to that at the bottom or top. Assuming the ratio is 4:1 based on the provided answer.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

The amplitude of a damped harmonic oscillator becomes halved in $\ minute$. After three minutes, the amplitude will becomes $\dfrac{1}{x}$ of initial amplitude, where $x$ is ?

  1. $8$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a damped oscillator, amplitude A(t) = A0 * exp(-bt/2m). Given A(1) = A0/2, then exp(-b/2m) = 1/2. After 3 minutes, A(3) = A0 * (exp(-b/2m))^3 = A0 * (1/2)^3 = A0/8. Thus x = 8.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

A particle performing SHM is found at its equilibrium at $  t=1\ sec$ and it is found to have a speed of $0.25 \mathrm{m} / \mathrm{s}  $ at $  \mathrm{t}=2\ \mathrm{sec}  $ . If the period of oscillation is $6\ \mathrm{sec}  $. Calculate amplitude of oscillation

  1. $ \frac{3}{2 \pi} \mathrm{m} $
  2. $ \frac{3}{ \pi} \mathrm{m} $
  3. $ \frac{6}{2 \pi} \mathrm{m} $
  4. $ \frac{6}{ \pi} \mathrm{m} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In SHM, x(t) = A sin(omega * t + phi). Equilibrium at t=1 means sin(omega + phi) = 0. Period T=6s, so omega = 2pi/6 = pi/3. At t=2, v = A * omega * cos(omega * t + phi) = 0.25. Solving these equations yields A = 3/(2pi).